Question

question 5 (12 marks)
suppose that ( n > 0 ) is an integer and ( a_0, dots, a_n ) are real numbers such that (\frac{a_0}{n + 1} + \frac{a_1}{n} + dots + \frac{a_{n - 1}}{2} + a_n = 0).
prove that the function ( f(x) = a_0x^n + a_1x^{n - 1} + dots + a_{n - 1}x + a_n ) has at least one root in ( (0, 1) ).

Explanation:

Step1: Define Auxiliary Function

Let \( F(x)=\frac{a_0}{n + 1}x^{n+1}+\frac{a_1}{n}x^{n}+\cdots+\frac{a_{n - 1}}{2}x^{2}+a_nx \). This function is constructed by integrating \( f(x) \) (since the derivative of \( F(x) \) with respect to \( x \) should be \( f(x) \), we check: \( F^\prime(x)=a_0x^n+a_1x^{n - 1}+\cdots+a_{n - 1}x + a_n=f(x) \)).

Step2: Evaluate \( F(0) \) and \( F(1) \)

• Evaluate \( F(0) \): Substitute \( x = 0 \) into \( F(x) \). All terms have a factor of \( x \) (except maybe the last term, but when \( x = 0 \), \( a_nx=0 \)), so \( F(0)=0 \).
• Evaluate \( F(1) \): Substitute \( x = 1 \) into \( F(x) \). We get \( F(1)=\frac{a_0}{n + 1}+\frac{a_1}{n}+\cdots+\frac{a_{n - 1}}{2}+a_n \). From the given condition, \( \frac{a_0}{n + 1}+\frac{a_1}{n}+\cdots+\frac{a_{n - 1}}{2}+a_n = 0 \), so \( F(1)=0 \).

Step3: Apply Rolle's Theorem

Rolle's Theorem states that if a function \( F(x) \) is continuous on \([a,b]\), differentiable on \((a,b)\), and \( F(a)=F(b) \), then there exists at least one \( c\in(a,b) \) such that \( F^\prime(c)=0 \).

Here, \( F(x) \) is a polynomial (so it is continuous on \([0,1]\) and differentiable on \((0,1)\)) and \( F(0)=F(1)=0 \). Therefore, by Rolle's Theorem, there exists some \( c\in(0,1) \) such that \( F^\prime(c)=0 \). But \( F^\prime(x)=f(x) \), so \( f(c)=0 \) for some \( c\in(0,1) \). This means \( f(x) \) has at least one root in \((0,1)\).

Answer:

The function \( f(x) \) has at least one root in \((0,1)\) (proven using Rolle's Theorem with the auxiliary function \( F(x) \)).